Prove $\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$

Question

Let $a,b,c$ be positive real numbers. Show that $$\sum_{cyc}\sqrt[3]{\dfrac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$$

I have tried C-S and Holder inequalities, without success. How to solve it?

Answer 1

The hint.

By Holder $$\left(\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b+c}}\right)^3\sum_{cyc}(b+c)(a^2+bc)^3(3a+4b+4c)^4\geq\left(\sum_{cyc}(a^2+bc)(3a+4b+4c)\right)^4.$$ Thus, it remains to prove that $$\left(\sum_{cyc}(a^2+bc)(3a+4b+4c)\right)^4\geq9(a+b+c)\sum_{cyc}(b+c)(a^2+bc)^3(3a+4b+4c)^4,$$ which is true by $uvw$ (this inequality is equivalent to $f(w^3)\geq0,$ where $f$ is a cocave function).

About $uvw$ see here: https://math.stackexchange.com/tags/uvw/info

https://artofproblemsolving.com/community/c6h278791

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.

Thus, we need to prove that $f(w^3)\geq0$, where $$f(w^3)=(27u^3-3uv^2-2w^3)^4+u(A(u,v^2)w^9+B(u,v^2)w^6+C(u,v^2)w^3+D(u,v^2)),$$ where $$A(u,v^2)=6042u^2-52v^2,$$ $$B(u,v^2)=-232389u^5+104382u^3v^2-5049uv^4,$$ $$C(u,v^2)=177147u^8+8748u^6v^2+78489u^4v^4-64881u^2v^6+459v^8$$ and $$D(u,v^2)=-1594323u^9v^2+1948617u^7v^4-618921u^5v^6-48114u^3v^8+5589uv^{10}.$$ Now, since $u\geq v\geq w$, we obtain: $$f''(w^3)=48(27u^3-3uv^2-2w^3)^2+6uA(uv^2)w^3+2uB(u,v^2)=$$ $$=48(27u^3-3uv^2-2w^3)^2+6u(6042u^2-52v^2)w^3+$$ $$+2u(-232389u^5+104382u^3v^2-5049uv^4)=$$ $$=-429786u^5+200988u^4v^2-9666u^2v^4+31068u^3w^3+264uv^2w^3+192w^6\leq0,$$ which says that $f$ is a concave function.

But a concave function gets a minimal value for an extreme value of $w^3$,

which says that it's enough to prove our inequality for an extreme value of $w^3$.

Now, $a$, $b$ and $c$ they are positive roots of the following equation. $$(x-a)(x-b)(x-c)=0$$ or $$x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=0$$ or $$w^3=x^3-3ux^2+3v^2x,$$ which says that the line $y=w^3$ and a graph of $f$,

where $f(x)=x^3-3ux^2+3v^2x$, have three common points (with maybe a degree of the point).

Now, $$f'(x)=3(x^2-2ux+v^2),$$ which says that $x_{max}=u-\sqrt{u^2-v^2}$ and $x_{min}=u+\sqrt{u^2-v^2}$ and we see that

$w^3$ gets a maximal value, when the line $y=w^3$ is a tangent line to the graph of $f$,

which happens for equality case of two variables.

Also, we see that $f$ gets a minimal value, when the line $y=w^3$ is a tangent line to the graph of $f$,

which happens for equality case of two variables or we need to check what happens for $w^3\rightarrow0^+$.

Id est, it's enough to prove that $f(w^3)\geq0$ in the following cases:

1. $b=c$.

Since this inequality is homogeneous we can assume $b=c=1$, which gives: $$((a^2+1)(3a+8)+2(a+1)(4a+7))^4\geq$$ $$\geq9(a+2)(2(a^2+1)^3(3a+8)^4+2(a+1)^4(4a+7)^4)$$ or $$(a-1)^2(81a^{10}+432a^9-1161a^8-10320a^7-2093a^6+115994a^5+361653a^4+525718a^3+437016a^2+184086a+364)\geq0,$$ which is true for all $a>0$;

2. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=1$.

Thus, we need to prove that: $$(a^2(3a+4)+(4a+3)+a(4a+4))^4\geq$$ $$\geq9(a+1)(a^6(3a+4)^4+a(4a+3)^4+(a+1)a^3(4a+4)^4)$$ or $$(a+1)^2(81a^{10}-27a^9-324a^8+87a^7+214a^6+29a^5+214a^4+87a^3-324a^2-27a+81)\geq0,$$ which is true.

Answer 2

Just a little hint

By AM GM $$\sum \sqrt[3] {\frac {a^2+bc}{b+c}}\ge 3\sqrt[3] {\prod \left (\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}$$

Now by Hölder's inequality $$3\sqrt[3] {\prod \left(\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}\ge 3\sqrt[3] {\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}+\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}}=3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}$$

Now again by AM GM on denominator

$$3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}\ge 3\sqrt[3] {\frac {3}{(a+b+c)}\sqrt[3] {a^2b^2c^2}}$$

Hope it helped any way